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已知点A(0,-2),B(0,4),动点P(x,y)满足
PA
PB
=y2-8
,则动点P的轨迹方程是 ___ .
人气:166 ℃ 时间:2020-04-06 08:36:17
解答
∵点A(0,-2),B(0,4),动点P(x,y)满足
PA
PB
=y2-8

则有(-x,-y-2)•(-x,4-y)=y2-8,即 x2+y2-2y-8=y2-8,
化简可得x2=2y,
故答案为:x2=2y.
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